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A vacuum flask (for keeping drinks hot) is modelled as a closed cylinder in which the internal radius is r cm and the internal height is h cm.

The volume of the flask is 1000 cm³. A flask is most efficient when the total internal surface area, A cm², is a minimum.

(i) Show that A = 2π2²+ 2000/2

(ii) Given that r can vary, find the value of r, correct to 1 decimal place, for which A has a stationary value and verify that the flask is most efficient when r takes this value.

User Kthakore
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1 Answer

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(i) To find the total internal surface area, we need to calculate the area of the curved surface and the areas of the top and bottom circles.

The curved surface area of the cylinder is given by the formula:

A_curved = 2πrh

The area of the top and bottom circles is given by the formula:

A_circles = 2πr²

The total internal surface area is the sum of the curved surface area and the areas of the top and bottom circles:

A = A_curved + A_circles

= 2πrh + 2πr²

Given that the volume of the flask is 1000 cm³, we have the equation:

πr²h = 1000

Solving for h, we get:

h = 1000 / (πr²)

Substituting this value of h into the equation for A, we have:

A = 2πrh + 2πr²

= 2πr(1000 / (πr²)) + 2πr²

= 2000/r + 2πr²

Simplifying further, we get:

A = 2000/r + 2πr²

= 2πr² + 2000/r

(ii) To find the value of r for which A has a stationary value, we differentiate A with respect to r and set the derivative equal to zero.

dA/dr = 4πr - 2000/r² = 0

Multiplying through by r², we get:

4πr³ - 2000 = 0

Solving for r, we have:

4πr³ = 2000

r³ = 2000 / (4π)

r = (2000 / (4π))^(1/3)

Calculating this value to one decimal place, we get:

r ≈ 3.4 cm

To verify that the flask is most efficient when r takes this value, we need to check the second derivative of A with respect to r.

d²A/dr² = 12πr² + 4000/r³

Substituting the value of r, we have:

d²A/dr² = 12π(3.4)² + 4000/(3.4)³

Calculating this value, we find that it is positive, indicating a minimum. Therefore, the flask is most efficient when r ≈ 3.4 cm.

User Acrollet
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